Calculating Reliability for Subsystems in Series

[normzone]

Thanks, [JeantheBigone] - that will come in very handy. I appreciate that very much.

 

[Miner]

To add a little more detail, there are two major types of systems from a reliability perspective. What Jean described is a SERIAL system. This is like to old style Christmas lights. When one light burned out, then entire string went dark. Any single element fails, and the entire system fails.

Another type is a PARALLEL system. This typically signifies redundancy, where if one element fails, there is a redundant element that takes over. The Apollo lunar lander had 3 redundant computers in case one failed. Usually all parallel elements must fail before the system fails. There are exceptions.

Then you can have a hybrid system which contain both serial and parallel elements. You can have imperfect redundancy also (e.g., run flat tires).

If you are interested in learning more about reliability, see Reliability Engineering Resources.

 

[normzone]

Thank you very much, [Miner]. I figured there would be another layer to the onion. Thanks for the homework links. I really appreciate all the information sharing people do here.

 

[JeantheBigone]

I was trying to think of a way to extend the tennis player analogy to include parallel systems. I didn't come up with anything that I liked.

 

[normzone]

I rely heavily on analogies for communicating ideas, but the tool only works so well, and then you wind up off the road and in the weeds or the ditch, depending on where you were driving

Perhaps something about managing a tennis tournament, and trying to prepare the trophies in advance ?

 

[JeantheBigone]

Let's try this.

Go back to Bob and Jane having a 50-50 chance of showing up to play tennis, but now you're only playing singles. So if either one or both show up, you get to play.

It's easy to figure out that now the chances of you getting at least ONE person to play tennis with are 75 %.

The formula is

R = 1 - [(1- R(Bob)) x (1 - R(Jane))]

where Bob's and Jane's reliability is still 0.5.

So

R = 1 - [(1- 0.5) x (1 - 0.5)]
= 1 - 0.25
= 0.75.

Just a thought......